A mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 30. cm3. If the combustion of this mixture releases 984 J of energy, to what volume (in L) will the gases expand against a constant pressure of 648 torr if all the energy of combustion is converted into work to push back the piston?

Answer:

300 K

Explanation:

First, we have find the specific heat capacity of the unknown substance.

The heat gained by the substance is given by the formula:

H = m*c*(T2 - T1)

Where m = mass of the substance

c = specific heat capacity

T2 = final temperature

T1 = initial temperature

From the question:

H = 200J

m = 4 kg

T1 = 200K

T2 = 240 K

Therefore:

200 = 4 * c * (240 - 200)

200 = 4 * c * 40

200 = 160 * c

c = 200/160

c = 1.25 J/kgK

The heat capacity of the substance is 1.25 J/kgK.

If 300 J of heat is added, the new heat becomes 500 J.

Hence, we need to find the final temperature, T2, when heat is 500 J.

Using the same formula:

500 = 4 * 1.25 * (T2 - 200)

500 = 5 * (T2 - 200)

100 = T2 - 200

=> T2 = 100 + 200 = 300 K

The new final temperature of the unknown substance is 300K.

Answer:

Explanation:

velocity of salmon with respect to water, v(s,w) = 5 mi/h at N 45° E

velocity of water with respect to ground, v(w,g) = 2 mi/h due east

Let the true velocity of salmon is velocity of salmon with respect to water is v(s,g)

First write the velocities in vector from

[tex]\overrightarrow{v}_{s,w}=5(Cos 45\widehat{i}+Sin 45\widehat{j})=3.54\widehat{i}+3.54\widehat{j}[/tex]

[tex]\overrightarrow{v}_{w,g}=2\widehat{i}[/tex]

Using the formula of relative speed,

[tex]\overrightarrow{v}_{s,w}=\overrightarrow{v}_{s,g}-\overrightarrow{v}_{w,g}[/tex]

[tex]3.54\widehat{i}+3.54\widehat{j}=\overrightarrow{v}_{s,g}-2\widehat{i}[/tex]

[tex]\overrightarrow{v}_{s,g}=5.54\widehat{i}+3.54\widehat{j}[/tex]

This i the true velocity of salmon.

The true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].

The given parameters;

The true velocity of the fish as a vector is calculated as follows;

[tex]v_f = v\ cos(\theta)i \ + \ v\ sin(\theta)j \ + \ 2i\\\\v_f = 5cos(45) i \ + 5sin(45)j \ + \ 2i\\\\v_f = 3.54i \ + \ 3.54j \ + 2i\\\\v_f = 5.54i \ + \ 3.54 j[/tex]

Thus, the true velocity of the fish as a vector is [tex]5.54i \ + \ 3.54j[/tex].

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Answer:

x₁ = 15 m, x₂ = 12 m , x_total = 27m, v₁ = 5 m / s ,  v₂ = - 3 m / s

Explanation:

In this exercise we will use the kinematics of uniform motion

        v = d / t

let's apply this equation for the first move

        v₁ = Δx / t = (x₂-x₀) / t

        v₁ = (12- (-3)) / 3

        v₁ = 5 m / s

the distance traveled is x₁ = 15 m

Now let's analyze the return movement

        v₂ = Δx / dt

        v₂ = (0 - 12) / 4

        v₂ = - 3 m / s

The negative sign indicates that the vehicle is moving to the left

the distance traveled is x₂ = 12 m

The total dystonia is

     x_total = x₁ + x₂

     x_total = 15 +12

     x_total = 27m

In the attached we have the graphics of the movement

Answer:

T = 74.3N

Explanation:

We are given;

v(t) = (2.00m/s²)t+(0.600m/s³)t²

So, when v = 9 m/s;

9 = 2t + 0.6t²

0.6t² + 2t - 9 = 0

Solving this quadratic equation,

t = -5.88 or 2.55

We'll pick t = 2.55 s

Now, kinematic acceleration will be the derivative of the acceleration.

Thus, a = dv/dt = 2 + 1.2t

So, acceleration at that time t = 2.55s is; a = 2 + 1.2(2.55) = 5.06 m/s²

Since the rope is subject to both acceleration and gravity, Tension is;

T = mg + ma

T = m(g + a)

T = 5(9.8 + 5.06)

T = 74.3N

Answer:

(a) 38.5m/s

(b) 64.4m/s

Explanation:

First, we can obtain the launch speed from the definition of kinetic energy:

[tex]K=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2K}{m}}\\\\[/tex]

Plugging in the given values, we obtain:

[tex]v=\sqrt{\frac{2(1550J)}{0.55kg}}\\\\v=75.0m/s[/tex]

Now, from the conservation of mechanical energy, considering the instant of launch and the instant of maximum height, we get:

[tex]E_0=E_f\\\\K_0=U_g_f+K_f\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0_x^2\\\\\frac{1}{2}mv_0^2=mgh_f+\frac{1}{2}mv_0^2\cos^2\theta\\\\\implies \cos\theta=\sqrt{1-\frac{2gh_f}{v_0^2}}[/tex]

And with the known values, we compute:

[tex]\cos\theta=\sqrt{1-\frac{2(9.8m/s^2)(140m)}{(75.0m/s)^2}}\\\\\cos\theta=0.513\\\\\theta=59.12\°[/tex]

Finally, to know the components of the launch velocity, we use trigonometry:

[tex]v_0_x=v_0\cos\theta=(75.0m/s)\cos(59.12\°)=38.5m/s\\\\v_0_y=v_0\sin\theta=(75.0m/s)\sin(59.12\°)=64.4m/s[/tex]

It means that the horizontal component of the launch velocity is 38.5m/s (a) and the vertical component is 64.4m/s (b).

Answer:

Asthenosphere

Explanation:

The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.

Answer:

C.

Explanation:

Answer:

Explanation:

Given that,

The speed of the car is

Vc = 97m/s

The radius of circular path of the car is

Rc = 420m

We want to find the angle of roadway banked β?

To determine the angle of roadway banked, we will use the formula

tanβ = Vc² / Rc•g

Where Vc = 97m/s, Rc = 420m and

g = 9.8m/s²

Then

tanβ = 97² / (420 × 9.8)

tanβ = 2.28596

β = ArcTan ( 2.28596)

β = 66.37°

The railway banked at an angle of 66.37°

Answer:

Banking angle is 66.35°

Explanation:

Given radius r=420m

Speed=97m/s

banking angle is A

Note before

(V)=√(r*gtanA)

√97=√420*9.81*tanA)taking square of both sides

97^2=420*9.81*tan A.

tanA=66.35°

A=66.35°

Answer:

1. 60 W

2. 24 W

3. 0.83 A

4. 25200 J

Explanation:

1. What we are simply asked to look for is Electrical Power. It is the rate electrical energy is being transferred.

It is given as:

[tex]P = IV[/tex]

where I = current and V = voltage.

Therefore, Power is:

P = 0.50 * 120 = 60 W

2. Power, as given in the formula above is:

P = 2.0 * 12

P = 24.0 W

3. According to the formula of Power, current is given as:

[tex]I[/tex] = [tex]\frac{P}{V}[/tex]

Power is 100 W and voltage is 120 V, therefore, current is:

[tex]I = \frac{100}{120} \\\\\\I = 0.83 A[/tex]

4. Recall that power is the time rate of transfer of electrical energy. Mathematically:

[tex]P = \frac{E}{t}[/tex]

where t = time

This means that Electrical energy is:

[tex]E = Pt[/tex]

Recall that Power is:

[tex]P = IV[/tex]

Therefore, Electrical energy is:

[tex]E = IVt[/tex]

[tex]E = 210 * 12 * 10\\\\\\E = 25200 J[/tex]

1. At "60 W" the bulb convert electric energy to light

2. Power used by lamp will be "24.0 W".

3. The current flow will be "0.83 A".

4. The electric energy delivered to the motor will be "25200 J".

According to the question,

1. Current, I = 0.50 A

  Voltage, V = 120 V

We know,

→ Power (P) = Current (I) × Voltage (V)

                     = 0.50 × 120

                     = 60 W

2. Current, I = 2.0 A

  Voltage, V = 12.0 V

then,

→ P = 2.0 × 12

       = 24.0 W

3. Power, P = 100 W

  Voltage, V = 120 V

We know,

→ Current, I = [tex]\frac{P}{V}[/tex]

                    = [tex]\frac{100}{120}[/tex]

                    = 0.83 A

4. Current, I = 210 A

Potential difference, V = 12.0  

We know,

→ E = Pt

    P = IV

or,

The electrical energy be:

→ E = IVt

       = 210 × 12 × 10

       = 25200 J

Thus the above answer is correct.

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Answer:

     v₀ = 240  m / s

Explanation:

This problem must be solved in two parts, first we must use the conservation of the moment, then the conservation of energy.

Let's start by applying moment conservation, to the system formed by the block and bullet, in this case the forces during the crash are internal and the moment is conserved

Instant starts. Before the crash

          p₀ = m v₀

Final moment. Right after the crash

         [tex]p_{f}[/tex] = (m + M) v

The moment is preserved

         Po =p_{f}

        M v₀ = (m + M) v

        v = m / (m + M) v₀        (1)

This is the speed with which the bullet block system comes out, now we can use energy conservation

Starting point. Right after the crash

         Em₀ = K = ½ (m + M) v²

Final point. Highest point of the path

         [tex]Em_{f}[/tex] = U = (m + M) g y

         Em₀ = Em_{f}

         ½ (m + M) v² = (m + M) g y

          v = √2 g y                 (2)

We substitute 1 in 2

              m / (m + M) v₀ = √ 2gy

            v₀ = (m + M) / m √ 2gy

Let's calculate

           v₀ = (0.0134 +0.800) /0.0134    √ (2 9.8 0.8)

           v₀ = 240  m / s

Answer:

298.04 m/s

Explanation:

Let m = mass of bullet = 0.0134 kg and M = mass of block = 0.800 kg.

Since the bullet becomes embedded in the block and rises a vertical height,h = 0.800 m

The kinetic energy change of mass + block =  potential energy of mass + block at height, h

ΔK = -ΔU

So, 1/2(m + M)(v² - V²) = -[(m + M)gh - 0] where v is the velocity of the bullet + block at height, h. Since the tension, T is the centripetal force at height, h, it follows that

T = (m + M)v²/r  r = length of cord = 1.44 m

v = √(Tr/(m + M)) = √4.76 N × 1.44 m/(0.800 + 0.0134)kg = √(6.8544/0.8134) = √8.427 = 2.9 m/s

So. 1/2(v² - V²) = -gh

v² - V² = -2gh

V = √(v² + 2gh) = √((2.9 m/s)² + 2 × 9.8 m/s² × 0.8 m) = √(8.41 + 15.68) = √24.09 = 4.91 m/s

This is the velocity of the bullet plus block at collision.

From the law of conservation of momentum,

momentum of bullet = momentum of bullet plus block

mv₀ = (m + M)V where v₀ = initial speed of bullet

v₀ = (m +M)V/m = (0.0134 kg + 0.800 kg)4.91 m/s ÷ 0.0134 kg = 3.994 ÷ 0.0134 kg = 298.04 m/s

Answer: 11.5 Hz, 29.83 m

Explanation:

Given

Linear density of the cable, μ = 0.2 kg/m

Tensión in the cables, T = 500 N

Wavelength of the wave, = 4.5 Waves

Distance between the poles, L = 20 m

Temperature of, t = 20° C

Speed of sound, v = 343 m/s

λ = length / number of waves =

λ = 20 / 4.5

λ = 4.44 m

Frequency of a standing wave is the same as frequency of a hum. Calculated using the formula

F = n/2L * √(T/μ)

F = 1/λ * √(T/μ)

F = 1/4.44 * √(500/0.2)

F = 0.23 * √2500

F = 0.23 * 50

F = 11.5 Hz

Wavelength of the hum,

λ = v/f

λ = 343 / 11.5

λ = 29.83 m

The wavelength of the hum produced by the cable is approximately 4.44 meters, and the frequency is approximately 11.26 Hz, calculated using the known linear density, tension, and distance between poles.

The student is asking about the properties of a standing wave created when the wind blows across a cable hung between two poles. Given the linear density (µ) of 0.2 kg/m, the tension in the cable (500 N), and the distance between poles (20 meters), we are to find the frequency and wavelength of the hum produced by the cable.

Firstly, we can calculate the wavelength (λ) of the wave using the information that 4.5 wavelengths fit between the two poles that are 20 meters apart:

λ = 20 meters / 4.5 wavelengths = 4.44 meters (approx)

Next, we can find the speed (v) of the wave on the cable using the formula:

v = sqrt(T / µ)

Plugging in the given values, we get:

v = sqrt(500 N / 0.2 kg/m) = sqrt(2500 m2/s2)

v = 50 m/s

Now, using the wave speed (v) and the wavelength (λ), we can calculate the frequency (f) of the wave using the formula:

f = v / λ

f = 50 m/s / 4.44 m = 11.26 Hz (approx)

Therefore, the wavelength is approximately 4.44 meters, and the frequency of the hum is approximately 11.26 Hz.

Answer:

P = 68.125 kW

P startup = 43.05 kW

Explanation:

The power  required to operate this ski lift is 43.05 KW.

The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units. A scalar quantity is power.

The lift is operating a a steady speed of 10 km/h = 10×(1000/3600) m/s = 25/9 m/s.

Average mass of the chair= 250 kg.

Acceleration of the chair = 25/9 /5 m/s² = 5/9 m/s².

Vertical component of this acceleration is = (5/9) ×{200/√(200²+1000²)} m/s²

= 0.1089 m/s²

Hence, the required height = 1/2 ×  0.1089 ×5² m = 1.362 m.

So, the total work done = mgh +1/2mv²

= 250×9.8×1.362 + 1/2× 250 × (25/9)² joule

= 215240.56 joule

So, the power  required to operate this ski lift =  215240.56 /5 watt = 43048.112 W = 43.05 KW.

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Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

Errors in an experiment can include improperly calibrated equipment, human error in measurement timing, and uncontrolled conditions. To reduce these errors, calibration, multiple trials, and consistent variables can be used. Recognizing whether the results are qualitative or quantitative helps in addressing the impacts of these errors.

When considering the question of what best describes an error the students made in the experiment, we can refer to common sources of experimental error. These can vary widely but often include issues with the methods used, errors with the apparatus, or uncontrolled conditions within the experiment. For instance, students may have used improperly calibrated equipment like an electronic balance, leading to systematic errors in measurement. Furthermore, the timing of events using devices like stopwatches can introduce human error if not stopped accurately.

An important thing to consider is whether these errors yield qualitative or quantitative results, as this will determine how we interpret the data. In the case of quantitative results, an inaccurate measurement has a direct numerical impact on the outcome, whereas qualitative errors might inform the observational or descriptive aspects of the results.

To reduce experimental error, students could implement controls such as calibrating equipment before use, running multiple trials to find an average, and ensuring that all variables aside from the independent variable are kept constant. By discussing potential experimental errors and how they might be mitigated, students can better understand how to refine their experimental design for more accurate and reliable results.

Answer:

in physics, a photon is a bundle of electromagnetic energy. It is the basic unit that makes up all light

Explanation:

Answer:

The quantum of light and other electromagnetic energy, regarded as a discrete particle having zero rest mass, no electric charge, and an indefinitely long lifetime. It is a gauge boson.

Answer:

Fraction of the specimen's is 0.4.

Explanation:

We know,

Mass = volume × density

Weigh= mass × g

          = volume × density× g

          = density× g × volume

          [tex]=\rho.g.V[/tex]

An object weighs less submerged due to buoyant force acting on it.

[tex]\therefore W_{wet}= W_{dry}-B[/tex]

[tex]B= W_{dry}-W_{wet}[/tex]

   [tex]=W_{\textrm{fluid displaced}}[/tex]

   [tex]=\rho_{fluid}. g.V_{submerged}[/tex]

Given that, the weighs of the specimen in dry is twice of the weighs in air.

[tex]W_{wet}=\frac 12W_{dry}[/tex]

Then ,

[tex]B= W_{dry}-W_{wet}[/tex]

   [tex]= W_{dry}-\frac12W_{dry}[/tex]

   [tex]=\frac12W_{dry}[/tex]

   [tex]=\rho_{Rock}. g.V_{Rock}[/tex]

Therefore,

[tex]\rho_{fluid}. g.V_{submerged}=\frac12\rho_{Rock}. g.V_{Rock}[/tex]

[tex]\Rightarrow \rho_{Rock}. g.V_{Rock}=2\rho_{fluid}. g.V_{submerged}[/tex]

[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}. g}{\rho_{Rock}.g}[/tex]

[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\rho_{fluid}}{\rho_{Rock}}[/tex]

[tex]\Rightarrow \frac{.V_{Rock}}{V_{submerged}}=\frac{2\times 1.0 \times 10^3\ kg /m^3}{5.0\times 10^3 \ kg/m^3}[/tex]

                     =0.4

Fraction of the specimen's is 0.4.

Tarzan's speed just before his feet touch the ground is 7.4 m/s, calculated using conservation of energy. The net change in internal energy as he bends his knees and stops is 2603.4 J, equivalent to the loss of kinetic energy.

Part (a): Speed of Tarzan Before Touchdown

To find the speed of Tarzan just before his feet touch the ground, we can use the principle of conservation of energy. Initially, Tarzan has gravitational potential energy due to being at a height of 2.8 m above the ground. When he lets go, this potential energy converts to kinetic energy as he falls. At the instant before his feet touch the ground, he is at a height of 0.8 m (2.8 m - 2 m), and the gravitational potential energy at this point converts into kinetic energy. Using the conservation of energy:

Initial Potential Energy = Final Kinetic Energy

mghinitial = 1/2*mv^2


Plugging in the values:
94 kg × 9.8 m/s2 × 2.8 m = (1/2) × 94 kg × v2

After solving the equation, we find:

v = 7.4 m/s

Part (b): Net Change in Internal Energy

The net change in internal energy is the difference between the kinetic energy just before the feet touch the ground and the energy when Tarzan is crouched. Assuming no losses, all the kinetic energy converts into internal energy. Without specific details on work done by Tarzan to flex his muscles or any other form of energy conversion, the kinetic energy that is no longer present in the motion would be assumed to convert fully to internal energy (such as heat). The net change in internal energy equates to the kinetic energy which is lost:

Change in Internal Energy = 1/2*mv^2 - 0 J, since he is at rest.

Using the speed found in part (a) and Tarzan's mass: 1/2*× 94 kg × (7.4 m/s)2

This calculation results in a net change in internal energy of: 2603.4 J

Answer:

a. 3π/2 b. 0.36 m c. 0.234 m/s d. 42.55 m/s²

Explanation:

Here is the complete question

A mass  m  =  1.1  kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant  k  =  130  N/m and negligible mass. The mass undergoes simple harmonic motion when placed in vertical motion, with its position given as a function of time by  y ( t ) =  A c o s ( ω t  −  ϕ ) , with the positive y-axis pointing upward. At time  t  =  0  the mass is observed to be passing through its equilibrium height with an upward speed of  v 0 =  3.9  m/s.

A. Find the smallest positive value of  ϕ ,  in radians.

B. Calculate the value of  A  in meters.  

C. What is the mass's velocity along the y-axis in meters per second, at the time  t  = 0.15  s?  

D. What is the magnitude of the mass's maximum acceleration, in meters per second squared?

Solution

a. Since y ( t ) =  A c o s ( ω t  −  ϕ ), the smallest possible value for ϕ is gotten when c o s ( ω t  −  ϕ ) = 0 ⇒ ω t  −  ϕ = cos⁻¹ 0 = π/2

ω t  −  ϕ = π/2.

At t = 0, ω t  −  ϕ = ω  0 −  ϕ = 0 −  ϕ = π/2

−  ϕ = π/2

ϕ = -'π/2

Since this is a negative angle, we add 2π to the right side.

So, ϕ = -'π/2 + 2π = 3π/2

ϕ = 3π/2

b. Since v = Aω = A√(k/m) where v = maximum velocity at time t = 0 = 3.9 m/s. A = amplitude, k = spring constant = 130 N/m and m = mass = 1.1 kg

A = v/√(k/m) = 3.9 m/s/√(130 N/m/1.1 kg) = 3.9/√118.18 = 3.9/10.87 = 0.36 m

c. To find its velocity, we differentiate y(t)

So, v = dy(t)/dt = dA c o s ( ω t  −  ϕ )/dt = -'ωAsin( ω t  −  ϕ ) = v₀sin( ω t  −  ϕ )

v = v₀sin( ω t  −  ϕ ) = v₀sin( ω t  −  ϕ)

Substituting the value of the variables,

v = 3.9sin( 10.87 t  −  3π/2)

At t = 0.15 s,

v = 3.9sin( 10.87 × 0.15  −  3π/2)

v = 3.9sin( 1.6305  −  4.7124)

v = -'3.9sin( -3.0819)

v = -'3.9 × - 0.06

v = 0.234 m/s

d. The maximum acceleration, a

a = Aω² = Ak/m = 0.36 × 130/1.1 = 42.55 m/s²

Answer:

a) F = 10.4 rad/s

b) A = 0.375 m

c) ϕ = 3π/2

d) V(t) = -ωAsin( ω t - 3π/2 )

e) V = 0.144 m/s

f) a = 40.625 m/s²

Explanation: Given that

mass m = 1.2 kg

The spring constant k = 130 N/m Time t = 0

Distance d = 0.35 m 

y( t ) =  A c o s ( ω t  −  ϕ )

At time  t  =  0

Speed of  Vo = 3.9  m/s.

a) Find the angular frequency of oscillations in radians per second

W = √(k/m)

2πF = √(k/m)

F = 1/2π√(k/m)

F = 1/2π √(130/1.2)

F = 1.66Hz

1 Henz = 6.28 rad/s therefore,

F = 1.66 × 6.28

F = 10.4 rad/s

b) Calculate the value of  A  in meters.  

V = Aω = A√(k/m)

where V = 3.9 m/s the maximum velocity at time t = 0

A = amplitude

A = v/√(k/m)

A = 3.9/√(130/1.2)

A = 3.9/10.4

A = 0.375 m

c. Determine the value of  ϕ in radians

If y( t ) =  A c o s ( ω t  −  ϕ ) We can obtain the smallest possible value of ϕ when c o s ( ω t  −  ϕ ) = 0

ω t  −  ϕ = cos⁻¹ 0 = π/2

ω t  −  ϕ = π/2.

At t = 0,

ω(0) −  ϕ = π/2

−  ϕ = π/2

ϕ = -'π/2

This is a negative angle, let us add 2π to the right side. So,

ϕ = -'π/2 + 2π = 3π/2

ϕ = 3π/2

d. Enter an expression for velocity along y axis as function of time in terms of A, ϕ and t using the value of ϕ from part c.

To find expression for velocity, we differentiate y(t) with respect to time t So,

V = dy/dt = dA c o s ( ω t  −  ϕ )/dt

V = -ωAsin( ω t  −  ϕ )

Therefore

V(t) = -ωAsin( ω t + π/2 )

Or

V(t) = -ωAsin( ω t - 3π/2 ) ...... (1)

e. What is the velocity of mass at time t = 0.25 s?

From equation (1)

V(t) = V₀sin( ω t  −  ϕ )

Substituting the value of the variables,

V = 3.9sin( 10.4t  −  3π/2)

At t = 0.25 s,

V = 3.9sin( 10.4 × 0.25 − 3π/2)

V = 3.9sin( 2.6 −  4.7124)

V= -3.9sin( -2.1124)

V= -3.9 × - 0.037

V = 0.144 m/s

f. What is the magnitude of mass's maximum acceleration?

The maximum acceleration a = Aω²

a = Aω² =

a = Ak/m

a = 0.375 × 130/1.2

a = 40.625 m/s²

Explanation:

Given that,

Diameter of the gold wire, d = 0.84 mm

Radius, r = 0.42 mm

Electric field in the wire, E = 0.49 V/m

(a) Electric current density is given by :

[tex]J=\dfrac{I}{A}[/tex]

And electric field is :

[tex]E=\rho J[/tex]

[tex]\rho[/tex] is resistivity of Gold wire

So,

[tex]E=\dfrac{\rhi I}{A}\\\\I=\dfrac{EA}{\rho}\\\\I=\dfrac{0.49\times \pi (0.42\times 10^{-3})^2}{2.44\times 10^{-8}}\\\\I=11.12\ A[/tex]

(b) The potential difference between two points in the wire is given by :

[tex]V=E\times l\\\\V=0.49 \times 6.4\\\\V=3.136\ V[/tex]

(c) Resistance of a wire is given by :

[tex]R=\rho \dfrac{l}{A}\\\\R=2.44\times 10^{-8}\times \dfrac{6.4}{\pi (0.42\times 10^{-3})^2}\\\\R=0.281\ \Omega[/tex]

Hence, this is the required solution.

Final answer:

To find the current carried by the gold wire, use Ohm's law. To find the potential difference between two points in the wire 6.4 m apart, multiply the electric field by the distance. To find the resistance of a 6.4-m length of the wire, use the formula R = ρL/A.

Explanation:

To find the current carried by the gold wire, we can use Ohm's law which states that the electric field (E) is equal to the product of the current (I) and the resistance (R) of the wire. Rearranging the equation, we can solve for the current: I = E/R. Given that the diameter of the wire is 0.84 mm, we can calculate the cross-sectional area using the formula for the area of a circle (A = πr^2). With the cross-sectional area, we can find the resistance by using the formula R = ρL/A, where ρ is the resistivity of gold and L is the length of the wire. Finally, we can substitute the values into the equation to find the current carried by the wire.

(b) To find the potential difference between two points in the wire 6.4 m apart, we can multiply the electric field by the distance between the two points: V = E × d.

(c) The resistance of a 6.4-m length of this wire can be found using the formula R = ρL/A, where ρ is the resistivity of gold, L is the length of the wire, and A is the cross-sectional area of the wire.

Answer:

Final speed = 148.21m.s

Time of flight = 7.82seconds

Explanation:

The motion of the body is a projectile motion.

Projectile is a motion created by an object launched in air and allowed to fall to freely under the influence of gravity.

Taking the maximums height reached H = 75m

Angle of launch = 15°

Using the maximum height formula to get the velocity U of the object

H = U²sin²theta/2g

Where g is the acceleration due to gravity = 9.81m/s²

75 = U²(sin15°)²/2(9.81)

1471.5 = U²(sin15°)²

1471.5 = 0.06699U²

U² = 1471.5/0.06699

U² = 21,965.9

U = √21,965.9

U = 148.21m/s

The time taken for the skier to reach the bottom of this hill starting from rest is the time of flight T.

T = 2Usintheta/g

T = 2(148.21)sin15°/9.81

T = 296.42sin15°/9.81

T = 76.72/9.81

T = 7.82seconds

Answer:

Sirius A is 1.608 times the size of the Sun.

Explanation:

The radiant flux establishes how much energy an observer or a detector can get from a luminous source per unit time and per unit surface area.

[tex]R_{p} = \frac{L}{4\pi r^2}[/tex]  (1)

Where [tex]R_{p}[/tex] is the radiant power received from the source, L is its intrinsic luminosity and r is the distance.

The Stefan-Boltzmann law is defined as:

[tex]R_{p} = \sigma \cdot T^{4}[/tex]  (2)

Where [tex]R_{p}[/tex] is the radiant power, [tex]\sigma[/tex] is the Stefan-Boltzmann constant and T is the temperature.

Then, equation 2 can be replaced in equation 1

[tex]\sigma \cdot T^{4} = \frac{L}{4\pi r^2}[/tex] (3)

Notice that L is the energy emitted per second by the source.

Therefore, r can be isolated from equation 3.

[tex] r^2 = \frac{L}{4\pi \sigma\cdot T^{4}}[/tex]

[tex] r = \sqrt{\frac{L}{4\pi \sigma\cdot T^{4}}}[/tex]  (4)

The luminosity of the Sun can be estimated isolating L from equation 3.

[tex]L = (4\pi r^2)(\sigma \cdot T^{4}) [/tex]

but, [tex]r = 696.34x10^{6}m[/tex] and [tex]T = 5800K[/tex]

[tex]L_{Sun} = 4\pi (696.34x10^{6}m)^2(5.67x10^{-8} W/m^{2} K^{4} )(5800K)^{4}) [/tex]

[tex]L = 3.90x10^{26} W[/tex]

To find the luminosity of Sirius A, the following can be used:

[tex]\frac{L_{SiriusA}}{L_{sun}} = 23[/tex]

[tex]{L_{SiriusA}} = (3.90x10^{26} W)(23)[/tex]

[tex]{L_{SiriusA}} = 8.97x10^{27}W[/tex]

Finally, equation 4 can be used to determine the radius of Sirius A.

[tex] r = \sqrt{\frac{8.97x10^{27}W}{4\pi (5.67x10^{-8} W/m^{2} K^{4})(10000K)^{4}}}[/tex]

[tex]r = 1.12x10^{9}m[/tex]

So, Sirius A has a radius of [tex]1.12x10^{9}m[/tex]

[tex]\frac{1.12x10^{9}m}{696.34x10^{6}m} = 1608[/tex]

Hence, Sirius A is 1.608 times the size of the Sun.

Answer:

Explanation:

What happens when a flame is deprived of oxygen? Carbon dioxide molecules are heavier than air. Because of this, they push the oxygen and other molecules in the air out of the way as they sink down over the flame and candle. When oxygen is pushed away from the wick, it can't react with the wax anymore. This makes the flame go out.

So if the flame is exposed to air, the flame will rise.

Answer: When gas is exposed to a small flame it explodes into a firerer blow torch at an unbelievable temperature of almost 450 degrees ^F

Explanation: This is just like igniting a lighter then placing an aerosol can of

fumeable gas then spraying the flammeable gas from the aerosol can into the flame of the lighter which will then give you a flamethrower.

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

V = 0.5 m/s

Explanation:

Given,

Mass of the raft,M= 180 Kg

initial velocity of raft,u = 0 m/s

Final velocity of the raft,V = ?

Mass of swimmer 1, m₁ = 50 Kg

initial velocity of swimmer, u₁ = 0 m/s

Final velocity of the swimmer 1, v₁ = 3 m/s

Mass of swimmer 2, m₂ = 80 Kg

Initial velocity of the swimmer 2, u₂ = 0 m/s

Final velocity of the swimmer 2, v₂ = - 3 m/s

Using conservation of momentum

M u + m₁ u₁ + m₂ u₂ = M V + m₁ v₁ + m₂ v₂

M x 0 + m₁ x 0₁ + m₂x 0 = 180 x V + 50 x 3 + 80 x (-3)

180 V = 90

V = 0.5 m/s

Hence, the speed if the raft is 0.5 m/s in the direction of swimmer 1.

Answer:

Energy is conserved.

Explanation:

According to law of conservation of energy ,energy can neither be created nor be destroyed .

When the player kicks the soccer ball, the kinetic energy is transferred to the ball and it will roll for a while and then stops. The combination of force and distance is called as work.Greater the force faster the ball will go.

The force with which the soccer player player kicked the ball will be having it effects on the ball. As the ball rolls in the ground  kinetic energy which the ball got from the player will be lost in the form of heat due to friction.

According to the Newton first law of motion,the object continues to remain  in rest or move with constant speed unless acted upon by the external force.

Question:

(a) She catches the Frisbee and holds it.  

Answer:

The correct option is;

A perfectly inelastic collision

Explanation:

A perfectly inelastic collision is one in which there is maximum amount of loss of kinetic energy in the system. In a perfectly inelastic collision, the colliding members lose their initial speed and they stick together resulting in a loss of kinetic energy.

Since she catches and holds on to the Frisbee, the kinetic energy of the Frisbee is lost as she holds on to it so as to combine her mass to that of the Frisbee.

The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction.

The final velocity of the truck driver does not change after making a U-turn and driving in the opposite direction. While the truck's speed changes, the velocity remains the same because velocity takes into account both the speed and direction of motion. Since the speed remains constant at 30 mph and the direction changes, the velocity is still 30 mph, just in the opposite direction.

Answer:

The cytoplasm holds all the organelles, except the nucleus.

Explanation:

The cytoplasm holds all organelles, with the primary exception being the nucleus, which is separated by a nuclear envelope. Other organelles like chloroplasts, mitochondria, and vacuoles, are held within the cytoplasm.

The cytoplasm, the watery fluid inside cells, houses almost all the organelles. The primary organelle that it does not encompass is the nucleus. The nucleus stays separated from the cytoplasm by a nuclear envelope.

Other organelles such as chloroplasts, mitochondria, and vacuoles are embedded within the cytoplasm. These organelles perform various roles within cells, while the nucleus controls the cell's activities and contains the cell's genetic information.

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Answer:

what is happiness

Explanation:

Answer:

Love is the answer to the question of how do you make someone do something that is not to their benefit. How do we stop bad people from doing bad things? How do we overcome greed? How can we find happiness if we always know that we have a limited time on earth?

Explanation: